Question

At time $t=0$, terminal $A$ in the circuit shown in the figure is connected to $B$ by a key and an alternating current $I(t)=$ $I_0\cos (\omega t)$, with $I_0=1$ A and $\omega =500rad/s$ starts flowing in it with the initial direction shown in the figure. At $t=7\pi /6\omega$, the key is switched from $B$ to $D$. Now onwards only $A$ and $D$ are connected. A total charge $Q$ flows from the battery to charge the capacitor fully. If $C=20\mu F,R=10\Omega$ and the battery is ideal with emf of $50V$, identify the correct statement(s).

• A. Magnitude of the maximum charge on the capacitor before $t=\frac{7\pi }{6\omega }$ is $1\times 10^{-3}C$
• B. The current in the left part of the circuit just before $t=\frac{7\pi }{6\omega }$ is clockwise.
• C. Immediately after $A$ is connected to $D$, the current in $R$ is $10A$.
• D. $Q=2\times 10^{-3}C$.

Answer 1

Answer: (C), (D)

If $q$ represents the charge on capacitor's upper plate:
$I(t)=I_0\cos (\omega t)=\frac{dq}{dt}$
$\Rightarrow q(t)=\frac{I_0}{\omega }\sin (\omega t)$
Max charge $=\frac{1A}{500rad{s}^{-1}}=2\times 10^{-3}C$
$I(t)=I_0\cos (\omega t)=\frac{dq}{dt}$
$\Rightarrow q(t)=\frac{I_0}{\omega }\sin (\omega t)$
Max charge $=\frac{1A}{500rad{s}^{-1}}=2\times 10^{-3}C$
Charge on upper plate at $t=\frac{7\pi }{6\omega }$
$=\frac{I_0}{\omega }\sin \left(\frac{7\pi }{6}\right)=-\frac{I_0}{2\omega }$
When capacitor is fully charged, charge on upper plate
$=50V\times 20\mu F=1\times 10^{-3}C$
$\therefore Q=1\times 10^{-2}C-\left(-\frac{I_0}{2\omega }\right)=2\times 10^{-3}C$
When capacitor is fully charged, charge on upper plate
$=50V\times 20\mu F=1\times 10^{-3}C$
$\therefore Q=1\times 10^{-2}C-\left(-\frac{I_0}{2\omega }\right)=2\times 10^{-3}C$
Voltage across capacitor when $A$ and $D$ are connected
$=\frac{1\times 10^{-3}C}{20\times 10^{-6}\mu F}=50V$
Total voltage across resistor $=100V$
$\Rightarrow$ current $=\frac{100V}{10\Omega }=10A$
Current is anticlockwise.